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Properties

Label 18T88
Order \(162\)
n \(18\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $C_3\wr C_3:C_2$

Related objects

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Group action invariants

Degree $n$ :  $18$
Transitive number $t$ :  $88$
Group :  $C_3\wr C_3:C_2$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,4,17)(2,3,18), (1,8,14)(2,7,13)(3,10,16)(4,9,15)(5,11,17)(6,12,18), (1,3)(2,4)(5,12)(6,11)(7,15)(8,16)(9,13)(10,14)(17,18)
$|\Aut(F/K)|$:  $6$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
6:  $S_3$ x 4
18:  $C_3^2:C_2$
54:  $(C_3^2:C_3):C_2$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 3: $S_3$

Degree 6: $S_3$

Degree 9: $(C_3^3:C_3):C_2$

Low degree siblings

9T21 x 3, 18T88 x 2

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy Classes

Cycle TypeSizeOrderRepresentative
$ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $6$ $3$ $(11,14,15)(12,13,16)$
$ 3, 3, 3, 3, 1, 1, 1, 1, 1, 1 $ $6$ $3$ $( 5, 8, 9)( 6, 7,10)(11,14,15)(12,13,16)$
$ 3, 3, 3, 3, 1, 1, 1, 1, 1, 1 $ $3$ $3$ $( 5, 8, 9)( 6, 7,10)(11,15,14)(12,16,13)$
$ 3, 3, 3, 3, 1, 1, 1, 1, 1, 1 $ $3$ $3$ $( 5, 9, 8)( 6,10, 7)(11,14,15)(12,13,16)$
$ 2, 2, 2, 2, 2, 2, 2, 2, 2 $ $27$ $2$ $( 1, 2)( 3,17)( 4,18)( 5,12)( 6,11)( 7,15)( 8,16)( 9,13)(10,14)$
$ 6, 6, 2, 2, 2 $ $27$ $6$ $( 1, 2)( 3,17)( 4,18)( 5,12, 8,16, 9,13)( 6,11, 7,15,10,14)$
$ 6, 6, 2, 2, 2 $ $27$ $6$ $( 1, 2)( 3,17)( 4,18)( 5,12, 9,13, 8,16)( 6,11,10,14, 7,15)$
$ 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1, 4,17)( 2, 3,18)( 5, 8, 9)( 6, 7,10)(11,14,15)(12,13,16)$
$ 3, 3, 3, 3, 3, 3 $ $6$ $3$ $( 1, 4,17)( 2, 3,18)( 5, 8, 9)( 6, 7,10)(11,15,14)(12,16,13)$
$ 3, 3, 3, 3, 3, 3 $ $18$ $3$ $( 1, 5,11)( 2, 6,12)( 3, 7,13)( 4, 8,14)( 9,15,17)(10,16,18)$
$ 9, 9 $ $18$ $9$ $( 1, 5,11, 4, 8,14,17, 9,15)( 2, 6,12, 3, 7,13,18,10,16)$
$ 9, 9 $ $18$ $9$ $( 1, 5,11,17, 9,15, 4, 8,14)( 2, 6,12,18,10,16, 3, 7,13)$

Group invariants

Order:  $162=2 \cdot 3^{4}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [162, 19]
Character table:    
      2  1  .  .  1  1  1   1   1  .  .  .  .  .
      3  4  3  3  3  3  1   1   1  4  3  2  2  2

        1a 3a 3b 3c 3d 2a  6a  6b 3e 3f 3g 9a 9b
     2P 1a 3a 3b 3d 3c 1a  3c  3d 3e 3f 3g 9a 9b
     3P 1a 1a 1a 1a 1a 2a  2a  2a 1a 1a 1a 3e 3e
     5P 1a 3a 3b 3d 3c 2a  6b  6a 3e 3f 3g 9a 9b
     7P 1a 3a 3b 3c 3d 2a  6a  6b 3e 3f 3g 9a 9b

X.1      1  1  1  1  1  1   1   1  1  1  1  1  1
X.2      1  1  1  1  1 -1  -1  -1  1  1  1  1  1
X.3      2  2  2  2  2  .   .   .  2  2 -1 -1 -1
X.4      2 -1 -1  2  2  .   .   .  2 -1  2 -1 -1
X.5      2 -1 -1  2  2  .   .   .  2 -1 -1 -1  2
X.6      2 -1 -1  2  2  .   .   .  2 -1 -1  2 -1
X.7      3  .  .  A /A -1   B  /B  3  .  .  .  .
X.8      3  .  . /A  A -1  /B   B  3  .  .  .  .
X.9      3  .  .  A /A  1  -B -/B  3  .  .  .  .
X.10     3  .  . /A  A  1 -/B  -B  3  .  .  .  .
X.11     6  . -3  .  .  .   .   . -3  3  .  .  .
X.12     6  3  .  .  .  .   .   . -3 -3  .  .  .
X.13     6 -3  3  .  .  .   .   . -3  .  .  .  .

A = 3*E(3)^2
  = (-3-3*Sqrt(-3))/2 = -3-3b3
B = -E(3)
  = (1-Sqrt(-3))/2 = -b3